Answer
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Hint: Basically the chemical compounds that does not contain carbon and hydrogen are known as inorganic compounds, but we can see the presence of hydrogen in some of the inorganic compounds such water molecule (\[{H_2}O\] ) and $HCl$ which is produced in our stomach.
Complete step by step answer:
Atoms or ions that have the same electronic configuration or same number of electrons are isoelectronic to each other.
Number of electrons in $N{O^ + }$ ion are $14(7 + 8 - 1)$.
Number of electrons in ${O_2}$ are $16(8 + 8)$.
So due to the difference in the number of electrons in both the species we can conclude that these are not isoelectronic species.
As we all know that boron always forms covalent bond in its compounds, the reason behind this statement is that it has very small size so it requires a lot of energy to form ${B^{ + 3}}$ ion and thus it is easy for boron to share the electrons and complete its octet, that’s why $B$ is always covalent in its compounds.
As we all know that $T{l^ + }$ is more stable than $T{l^{ + 3}}$ ion, this happens due to inert pair effect, as we move down in a group the stability of lower oxidation states increases and the stability of higher oxidation states increases, hence the $T{l^ + }$ ion is much more stable than $T{l^{ + 3}}$ ion.
$LiAl{H_4}$ is also known as lithium aluminium hydride. It is a very powerful versatile reducing agent in organic chemistry. It is an unselective reducing agent for polar double bonds and also it is a source of hydride ion ${H^ - }$ .
Hence, among all of these statements, statement (A) is not true, $N{O^ + }$ is isoelectronic with ${O_2}$.
Note:
The inert pair effect is the tendency of electrons present in the outermost s-orbital of an atom to remain unshared in compounds. This is basically applicable for only two electrons present in the outermost atomic s-orbital. These two electrons have less tendency to participate in bond formation.
Complete step by step answer:
Atoms or ions that have the same electronic configuration or same number of electrons are isoelectronic to each other.
Number of electrons in $N{O^ + }$ ion are $14(7 + 8 - 1)$.
Number of electrons in ${O_2}$ are $16(8 + 8)$.
So due to the difference in the number of electrons in both the species we can conclude that these are not isoelectronic species.
As we all know that boron always forms covalent bond in its compounds, the reason behind this statement is that it has very small size so it requires a lot of energy to form ${B^{ + 3}}$ ion and thus it is easy for boron to share the electrons and complete its octet, that’s why $B$ is always covalent in its compounds.
As we all know that $T{l^ + }$ is more stable than $T{l^{ + 3}}$ ion, this happens due to inert pair effect, as we move down in a group the stability of lower oxidation states increases and the stability of higher oxidation states increases, hence the $T{l^ + }$ ion is much more stable than $T{l^{ + 3}}$ ion.
$LiAl{H_4}$ is also known as lithium aluminium hydride. It is a very powerful versatile reducing agent in organic chemistry. It is an unselective reducing agent for polar double bonds and also it is a source of hydride ion ${H^ - }$ .
Hence, among all of these statements, statement (A) is not true, $N{O^ + }$ is isoelectronic with ${O_2}$.
Note:
The inert pair effect is the tendency of electrons present in the outermost s-orbital of an atom to remain unshared in compounds. This is basically applicable for only two electrons present in the outermost atomic s-orbital. These two electrons have less tendency to participate in bond formation.
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